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Confusing $\sqrt{2}$ is irrational proof

Let $a, b$ be arbitrary positive integers.

Then $2b^{2}$ is divisible by 2 an odd number of times, while $a^{2}$ is divisible by 2 an even number of

times, so $2b^{2} \neq a^{2}$ and therefore $|2b^{2}a^{2}| \geqq 1.$

If $\frac{a}{b} \gt 3- \sqrt{2}$ then $\frac{a}{b} \gt 3-2\sqrt{2}\gt\frac{1}{6}$

If $\frac{a}{b} \leqq 3-\sqrt{2}$ then

$|\sqrt{2} - \frac{a}{b}| = \frac{|2b^{2}-a^{2}|}{b^{2}(\sqrt{2}+\frac{a}{b})} \geqq \frac{1}{b^2(\sqrt{2}+\frac{a}{b})} \geqq \frac{1}{3b^{2}}$

I get that $2b^{2} \neq a^{2}$ because $\sqrt{2} \neq \frac{a}{b}$ I also get that $|2b^{2}a^{2}|\geq1$ because were letting a, b be arbitrary positive integers.

I don't understand where the divisible by 2 an odd and even number of times comes from.

I notice that 3 is the smallest integer such that $3-\sqrt{2} $ and $3-2\sqrt{2} $return positive values.

Christian Blatter
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Andrew
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  • This is supposed to be a proof of what? – José Carlos Santos Nov 09 '19 at 15:31
  • My Bad. I edited the post. – Andrew Nov 09 '19 at 15:34
  • $\sqrt{2}=a/b$, hence $2b^2=a^2$, where $a$ and $b$ are coprime. Since $a^2$ is even, being $2b^2$, we conclude that $a$ is even, hence $a^2$ is divisible by $4$. Now finish with this post and forget the other "confusing irrational proof". – Dietrich Burde Nov 09 '19 at 15:35
  • And I guess that $a$ and $b$ are not arbitrary at all. Instead, I suspect that $\frac ab=\sqrt2$. – José Carlos Santos Nov 09 '19 at 15:36
  • "I get that $2b^2\neq a^2$ because $\sqrt 2 \neq \frac{a}{b}$" no it's not because of that. That one is the one you want to prove. @Jose Carlos Santos Seems like no. The writer just want to show that $|√2 - a/b| \geq 1/3b^2$ for any positive integers $a, b$, which implies equality cannot hold. – Azlif Nov 09 '19 at 16:01

1 Answers1

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The "divisible by $2$ an odd and even number of times" comes from the fact that if $a$ is divisible by $2$ $n$ times (i.e. $a=2^n \cdot m,m\in\mathbb{Z}),$ then $a^2$ is divisible by $2^{2n},$ and so it is divisible an even number of times. Likewise, $b^2$ is divisible by $2$ and even number of times, so $2b^2$ is divisible by $2$ an odd number of times.

Also, the inequality for $|2b^2a^2|$ could be simplified to $>$ since $a$ and $b$ are positive integers and the smallest possible value of $|2b^2a^2|$ is $2.$

You didn't say that you don't understand the last step so I assume you do (it's just using previous inequalities and the difference of squares formula).